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11:56 AM 31% webassign net My Notes o Ask Your Teacher 10 0/4 points I Previous Answers ColFunPhys1 17.P.021 A 1.96 x 10 C charge has coordinates X 0, y -2.00; a 3.27 x 10-9 c charge has coordinates x 3.00, y o; and a 5.50 x 10 C charge has coordinates x 3.00, y 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin (a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis) magnitude 34478 NIC direction (b) Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis) magnitude direction Additional Materials La Section 17.3 Submit Assignment Save Assignment Progress Home My Assignments Extension Request

Explanation / Answer

The field due to the 1st charge is in the +y direction

F1 = k*q1/r1^2 = 9.0x10^9*1.96x10^-9/0.02^2 = 44100N/C (+y)

F2 is in the -x direction

F2 = k*q2/r2^2 = 9.0x10^9*3.27x10^-9/0.030^2 = 32700N/C (-x)

F3 points in the -x & -y directions

F3 = k*q3/r3^2 = 9.0x10^9*5.50x10^-9/0.050^2 = 19800N/C
For this point cos() = -0.60 and sin() = -0.80
F3x = 19800*(-0.60) = -11880N/C
F3y = 19800*(-0.80) = -15840N/C

So Fx = -32700-15840 = -44580N/C

Fy = 44100 - 14976 = 29124N/C

So F = sqrt(44580^2 + 29124^2) = 5.40x10^4N/C

= arctan(29124/-44580) = -33.156 but = -32.156 + 180 = 147o (since Fx is negative)


b) a = F/m = E*q/m = 5.40x10^4*1.60x10^-19/1.67x10^-27 = 4.98x10^12m/s^2

the angle is the same as in a 147o

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